GALOIS THEORY BRIAN OSSERMAN

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1 GALOIS THEORY BRIAN OSSERMAN Galois theory relates the theory of field extensions to the theory of groups. It provides a powerful tool for studying field extensions, and consequently, solutions to polynomial equations. Much of our presentation is based on the book Topics in Algebra, by Herstein. 1. Automorphisms of field extensions Definition 1.1. Let K/F be a field extension. An automorphism of K/F is an isomorphism ϕ : K K such that ϕ(α) = α for all α F. Note that the definition implies in particular that ϕ will be a linear map of F -vector spaces. We can put an operation on the set of automorphisms of K/F by composing them. It is straightforward to check that this turns the set of automorphisms into a group (in fact, a subgroup of the group Sym(K) of all bijections from K to itself). Notation 1.2. Given a field extension K/F, let Aut(K/F ) denote the group of automorphisms of K/F. A basic consequence of the definition is: Proposition 1.3. Let ϕ be an automorphism of a field extension K/F, and f(x) F [x]. Let α 1,..., α n be the roots of f(x) lying in K. Then ϕ permutes the set {α 1,..., α n }. If also the set of α i generate K over F, then two automorphisms ϕ 1, ϕ 2 of K/F which agree on all the α i are equal. Thus, in this case we have an inclusion of Aut(K/F ) as a subgroup of Sym({α 1,..., α n }) = S n. Proof. For the first part, it suffices to observe that if α K is a root of f(x), then because ϕ fixes F and is a homomorphism, ϕ(α) is also a root of f(x). For the second part, because every element of K can be written as a rational function in the α i s, with coefficients in F, if both ϕ 1 and ϕ 2 keep F fixed and have the same values on all the α i, we conclude that they agree on all of K. Thus, we can think of Aut(K/F ) in terms of permuting roots of polynomials, and one basic question of Galois theory is which permutations of a given set of roots actually arise in this way. Example 1.4. Consider F = Q, and K = Q( 3 2). Then we know that [K : F ] = 3, and K contains exactly one root of the irreducible polynomial x 3 2 Q[x]. It follows from the second part of the proposition that Aut(K/F ) contains only the identity map in this case. Example 1.5. Consider F = Q, and K = Q( d), where d Z is squarefree and not equal to 0 or 1. Then [K : F ] = 2, and K contains two roots of the irreducible polynomial x 2 d, which we can write as ± d. We have the identity automorphism of K/F, keeping these roots fixed, but we also have the automorphism sending a + b d to a b d. This exchanges the roots ± d. According to the second part of the proposition, the size of Aut(K/F ) is at most S 2 = 2, so we conclude that we have found all of Aut(K/F ), which is necessarily the cyclic group of order 2. 1

2 The proposition says that if f(x) F [x] is irreducible of degree d, and we take a splitting field K/F of f(x), then the size of Aut(K/F ) is at most d!. We happen to know that this is also the maximum possible degree of K/F, so it is natural to wonder how these are related. The fundamental theorem is the following. Theorem 1.6. Let K/F be a finite extension. Then Aut(K/F ) [K : F ]. To prove the theorem, we will first prove the following: Lemma 1.7. Let K be a field, and ϕ 1,..., ϕ n distinct automorphisms of K. Given α 1,..., α n K, if α 1 ϕ 1 (β) + + α n ϕ n (β) = 0 for all β K, then all the α i are 0. Proof. The proof is by contradiction. If there exists some choice of α 1,..., α n as in the statement, and not all 0, then among all such choices, we could choose such a set of α i with a minimal number of them nonzero. Let m be the number of nonzero α i. We will show that we cannot have m = 1, but that if m > 1, we can always construct a new choice with strictly fewer nonzero α i, contradicting minimality. First observe that if we choose β = 1, we have we cannot have m = 1, since then α 1 ϕ 1 (1) + + α n ϕ n (1) = α α n 1 = α α n 0. On the other hand, if m > 1, then without loss of generality we may assume that α 1, α n are nonzero. By hypothesis, ϕ 1, ϕ n are distinct automorphisms of K, so there exists γ K such that ϕ 1 (γ) ϕ n (γ). For every β K, multiplying the assumed relation by ϕ n (γ) we get but since βγ K too, we have also α 1 ϕ n (γ)ϕ 1 (β) + + α n ϕ n (γ)ϕ n (β) = 0, 0 = α 1 ϕ 1 (γβ) + + α n ϕ n (γβ) = α 1 ϕ 1 (γ)ϕ 1 (β) + + α n ϕ n (γ)ϕ n (β), and subtracting the two equations gives α 1 (ϕ 1 (γ) ϕ n (γ))ϕ 1 (β) + α 2 (ϕ 2 (γ) ϕ n (γ))ϕ 2 (β) +... α n 1 (ϕ n 1 (γ) ϕ n (γ))ϕ n 1 (β) = 0 for all β K. But by hypothesis, α 1 (ϕ 1 (γ) ϕ n (γ)) 0, and for each i if we had α i = 0, then also α i (ϕ i (γ) ϕ n (γ)) = 0, so we have obtained a nontrivial linear relation with strictly fewer than m nonzero coefficients, contradicting the minimality hypothesis. The proof of the theorem then brings a little linear algebra into the picture. Proof of Theorem 1.6. Let n = [K : F ], and suppose we had distinct ϕ 1,..., ϕ n+1 Aut(K/F ). Let β 1,..., β n be a basis of K over F. Then we consider the system of n linear equations in n + 1 variables ϕ 1 (β 1 )x ϕ n+1 (β 1 )x n+1 = 0 ϕ 1 (β n )x ϕ n+1 (β n )x n+1 = 0. Note that the coefficients of this system are in K, not in F. By linear algebra, there exist α 1,..., α n+1 K such that setting all the x i equal to α i gives solutions to the above system of equations. We claim that this implies that in fact, we have ϕ 1 (β)α ϕ n+1 (β)α n+1 = 0 2.

3 for all β K. Indeed, since the β i form a basis of K over F, we can write β = i γ iβ i for some γ i F. Then ( ) ( ) ϕ 1 (β)α ϕ n+1 (β)α n+1 = ϕ 1 γ i β i α ϕ n+1 γ i β i α n+1 i i = i γ i (ϕ 1 (β i )α ϕ n+1 (β i )α n+1 ) = 0. where we have used above that ϕ j (γ i ) = γ i for all i, j by hypothesis. But this then contradicts Lemma 1.7. Example 1.8. For q = p r, consider the extension F q /F p, which we know has degree r. The Frobenius map ϕ sending α to α p is a homomorphism from F q to itself which fixes F p, and which you prove in your homework is bijective, so it is an automorphism of F q over F p. We claim that ϕ has order r. If we compose ϕ with itself s times, we get the map sending α to α ps. We know that α pr = α for all α F q, so if we choose s = r, we get the identity map on F q. Thus, ϕ has order at most r. On the other hand, if s < r, the elements of F q which are sent to themselves consist of the subfield F p s, so we do not get the identity map if s < r. We conclude that ϕ has order r, as claimed. This means that ϕ generates a cyclic subgroup of Aut(F q /F p ) of order r. However, Theorem 1.6 says that Aut(F q /F p ) r, so we conclude that in fact, Aut(F q /F p ) is cyclic of order r, generated by ϕ. 2. Galois theory: statements As previously mentioned, Galois theory relates the study of field extensions to group theory. More specifically, we define: Definition 2.1. Given a field extension K/F, an intermediate field is a field L contained in K and containing F. We say an intermediate field is proper if it is not equal to K or F. Then Galois theory will in particular let us study intermediate fields of given extensions in terms of (sub)groups of automorphisms. Intermediate fields might seem rather abstract, but to see why this is useful: Example 2.2. Suppose α C has degree 4 over Q, so that it is a root of a degree-4 irreducible polynomial in Q[x]. Then Q(α) has degree 4 over Q, and one could ask whether or not there is a proper intermediate field. Why would this matter? Suppose L is such an intermediate field. Then [L : Q] must divide 4, and since L Q and L Q(α), we cannot have [L : Q] equal to 1 or 4, so we conclude that [L : Q] is 2. By Proposition , we conclude that L = Q( d) for some d Q, and also Q(α) = L( β) for some β L. Thus, every element of Q(α) (including α itself) can be expressed with repeated b + c d for some a, b, c, d Q. square roots, and more explicitly, can be written as a + d + This is a nontrivial conclusion it turns out that not every root of a degree-4 polynomial over Q can be written in this form. Theorem 1.6 motivates the following definition: Definition 2.3. A finite extension K/F is Galois if Aut(K/F ) = [K : F ]. In this case, we also write Gal(K/F ) = Aut(K/F ), and call it the Galois group of K over F. Before stating the main theorems of Galois theory, we explore an extended example. 3

4 Example 2.4. Let F = Q, and K = Q( 2, 3). Then we know that [K : F ] = 4, and the elements 1, 2, 3, 6 form a basis of K over F. According to Proposition 1.3, if ϕ Aut(K/F ), then ϕ( 2) = ± 2, and ϕ( 3) = ± 3, and moreover ϕ is determined by what it does to 2 and 3 (note that ϕ(1) = 1 automatically, and ϕ( 6) = ϕ( 2)ϕ( 3), so we can see this directly in this case). In this example, it is straightforward to check that there are in fact automorphisms allowing for any combination of ϕ( 2) and ϕ( 3), so we have produced 4 automorphisms of K/F in this way, and we see that K/F is Galois. Each automorphism has order 2, and it is straightforward to see that the automorphisms commute in this case, so we have that Gal(K/F ) = C 2 C 2. Observe that the subgroups of Gal(K/F ) consist of the trivial group, the whole group, and three subgroups of order 2, with one generated by each non-identity element. We now consider intermediate fields of K/F. Since [K : F ] = 4, any intermediate field L must be either K or F, or must have degree 2 over F. In this last case, we can write L = Q( d) for some d see Proposition of Artin. Now, we can write an arbitrary α K as α = a 1 + a a3 3 + a4 6 for a1,..., a 4 Q, and a brute force calculation shows that if α 2 Q, then at most one of the a i can be nonzero. Thus, we conclude that the only intermediate subfields of K/F are Q( 2), Q( 3), and Q( 6). In conclusion, we have a picture where the subgroup structure of Gal(K/F ) looks the same as the intermediate field structure of the extension K/F itself. To each intermediate field L, we can obtain a subgroup of Gal(K/F ) by considering Gal(K/L). Conversely, given a subgroup H Gal(K/F ), we can find a field L by considering all the elements α K such ϕ(α) = α for all α H. For instance, if ϕ is the element of Gal(K/F ) with ϕ( 2) = 2, ϕ( 3) = 3 and ϕ( 6) = 6, we see that the corresponding subfield is the field generated by 3, since only combinations of 1 and 3 are fixed by ϕ. In this way, we can go back and forth between subgroups of Gal(K/F ) and intermediate fields between F and K. The construction in the example generalizes to the following definition: Definition 2.5. Let K be a field, and H Aut(K) a subgroup of automorphisms of K. Then the fixed field of H is defined by K H := {α K : ϕ(α) = α α H}. It is straightforward to check that the fixed field is in fact a subfield. Example 2.6. Suppose that K/F is an extension, and H Aut(K/F ). Then from the definitions, we see that K H K and K H F, so K H is an intermediate field. The main theorem of Galois theory is the following: Theorem 2.7. Let K/F be a Galois extension. correspondence Then we have an inclusion-reversing bijective {Intermediate fields L of K/F } {Subgroups H Gal(K/F )} given by L Gal(K/L) K H H. Moreover, under this correspondence, for each intermediate field L the extension L/F is Galois if and only if the corresponding subgroup Gal(K/L) Gal(K/F ) is normal. Implicit in the theorem statement is that if K/F is Galois and L is an intermediate field, then K/L is also Galois. This is an immediate consequence of the following complementary theorem describing Galois extensions: 4

5 Theorem 2.8. Let K/F be a finite extension. Then the following are equivalent: (a) K/F is Galois; (b) K is the splitting field of some polynomial f(x) F [x] which does not have multiple roots in K; (c) For every f(x) F [x] irreducible such that there exists a root α K of f(x), then f(x) splits completely in K and does not have multiple roots. In particular, if F has characteristic 0, then K/F is Galois if and only if K is the splitting field of some polynomial f(x) F [x]. Corollary 2.9. If K/F is Galois, and L is an intermediate subfield, then K/L is also Galois. Proof. By one direction of Theorem 2.8, we have that K/F is the splitting field of a polynomial f(x) F [x] without any multiple roots in K. Then considering f(x) L[x], we have that K/L is still the splitting field of f(x), so by the converse of the theorem it is still Galois. 3. Galois theory: proofs Our first task is to prove Theorem 2.8, characterizing Galois extensions. Some of the ingredients of the proof are important as well for the proof of Theorem 2.7. The main ingredients of the proof involve a careful analysis of the properties of splitting fields. However, we begin with a basic aside. We already know that a homomorphism between fields is always injective, because the kernel would be a non-unit ideal. For finite extensions, we have a nice complement, as follows. Proposition 3.1. Let K/F be a finite field extension, and ϕ : K K a homomorphism such that ϕ(α) = α for all α F. Then ϕ is an automorphism of K/F. Proof. We already know that ϕ is injective, so it suffices to prove surjectivity. But an injection linear map from a finite-dimensional F -vector space to itself is necessarily surjective, so we are done. We will need a few more basic properties of splitting fields, which we list below: Proposition 3.2. Let K/F be a finite extension. Then: (i) Given a polynomial f(x) F [x], there exists a polynomial g(x) F [x] which is a multiple of f(x) and such that if L/K is the splitting field of g(x) over K, then L/F is also the splitting field of g(x) over F. (ii) Given a polynomial g(x) F [x], and splitting fields L/K and E/F of g(x) over K and F respectively, there exists an inclusion of E as a subfield of L which is the identity on F. Proof. (i) Let β 1,..., β n be a set of generators of K/F, and let f 1 (x),..., f n (x) be their minimal polynomials. Then let g(x) = f(x) f 1 (x) f n (x), and let L be the splitting field of g(x) over K. Then g(x) splits completely in L, and because its roots contain a generating set of K over F, we see that L is also a splitting field of g(x) over F. (ii) By definition, g(x) splits completely in L, so let α 1,..., α n be the roots. Then F (α 1,..., α n ) L is a splitting field of g(x) over F. By our theorem on the uniqueness of splitting fields, there is an isomorphism E F (α 1,..., α n ) which is the identity on F, giving the desired inclusion of E as a subfield of L. We now use basic properties of field automorphisms to a rather remarkable fact (the (a) implies (b) part of the below) which is not a priori related to automorphisms. Theorem 3.3. Let K/F be a finite field extension. Then the following are equivalent: (a) K/F is a splitting field of some polynomial in F [x]. 5

6 (b) For every irreducible polynomial f(x) F [x] such that f(x) has a root in K, then in fact f(x) splits completely in K. (c) For every finite field extension L/K, and every automorphism ϕ of L/F, the image of K under ϕ is contained in K. Proof. First, we see that (b) implies (a): let β 1,..., β n be a set of generators of K/F, and let f 1 (x),..., f n (x) be their minimal polynomials. Then let g(x) be the product of the f i (x): we have by hypothesis that each f i (x) splits completely in K, and hence g(x) splits completely as well. But the roots of g(x) generate K over F by construction, so K is a splitting field for g(x) over F. We next show that (a) implies (c). Choose f(x) F [x] so that K/F is a splitting field for f(x), and let L/K be any extension, and ϕ any automorphism of L/F. Let α 1,..., α n be the roots of f(x) in K. Because f(x) splits completely in K, these are also the roots of f(x) in L, so by Proposition 1.3, they are permuted by ϕ. Recalling that by definition, K = F (α 1,..., α n ), we see that ϕ(k) K, as desired. We now verify that (c) implies (b). Given an irreducible f(x) F [x], suppose α K is a root of f(x). As in Proposition 3.2 (i), let g(x) F [x] be a multiple of f(x) such that the splitting field L/K of g(x) over K also gives a splitting field L/F over F. Now, because f(x) divides g(x), we will also have that f(x) splits completely in L, so let α 1,..., α n be the roots. Set E = F (α 1,..., α n ) L, so that E/F is a splitting field of f(x). Then by our theorem on splitting fields, for any i there exists an automorphism ϕ of E/F sending α to α i. Applying the theorem again, this automorphism can be lifted to an automorphism ψ of L/F with the property that ψ(α) = α i. Now, α K, so by hypothesis, α i = ψ(α) K as well. We thus conclude that f(x) splits completely in K, as desired. Note that the proof actually shows that (c) holds also without the finiteness hypothesis. Corollary 3.4. Suppose we have a finite extension K/F. Then K/F is the splitting field of a polynomial in F [x] if and only if the following condition is satisfied: there exists a field extension L/K such that L/F is a splitting field of some polynomial in F [x], and for every automorphism ϕ of L/F, the image of K under ϕ is contained in K. Proof. First suppose that K/F is the splitting field of a polynomial in F [x]. Then by Proposition 3.2 (i), there exists an extension L/K such that L/F is a splitting field, and by Theorem 3.3, for every ϕ Aut(L/F ) we must have ϕ(k) K, as desired. Conversely, suppose we are given L/K such that L/F is a splitting field of some f(x) F [x], and such that ϕ(k) K for all ϕ Aut(L/F ). Given any extension E/K, and ψ Aut(E/F ), we claim that ψ(k) K. Let E /E be the splitting field of f(x). By our theorem on splitting fields, there is an automorphism ψ of E which restricts to ψ on E: in particular, we have ψ Aut(E /F ). Now, by Proposition 3.2 (ii), we know that we can imbed L as a subfield of E, via an inclusion which is the identity on K. Because L/F is a splitting field, Theorem 3.3 tells us that ψ (L) L, so by Proposition 3.1 we see that ψ gives an element of Aut(L/F ). By hypothesis, we then have ψ (K) K, and since ψ restricts to ψ on elements of E, and K E, we conclude ψ(k) K. This proves the claim, and we can then apply Theorem 3.3 to conclude that K/F is a splitting field. In order to characterize extensions, it is helpful to have a basic understanding of how the number of automorphisms of an extension relates to an intermediate field. Proposition 3.5. Let K/F be a finite extension, and let L be an intermediate field. Then Aut(K/F ) / Aut(K/L) is at most equal to the number of field inclusions ϕ : L K such that ϕ(α) = α for all α F. If further K/L is a splitting field of some polynomial, then we have equality. 6

7 Proof. Let S be the set of field inclusions ϕ : L K as in the statement, so that we are trying to prove that Aut(K/F ) = Aut(K/L) S. Given ψ Aut(K/F ), we obtain an element of S by restricting ψ to L. If ψ 1, ψ 2 Aut(K/F ) restrict to the same element of S, then ψ 1 ψ 1 2 is an automorphism of K which fixes L. Conversely, given ψ Aut(K/F ) and ϕ Aut(K/L), then ψ ϕ is an element of Aut(K/F ) which has the same restriction to L as ψ. Thus, we have constructed a function Aut(K/F ) S such that for each element of S, the set of ψ Aut(K/F ) mapping to that element is in bijection with Aut(K/L). This proves the first statement, and for the second it is then enough to see that when K/L is a splitting field, our map to S is surjective. But given ϕ S, observe that if K is a splitting field of f(x) L[x] over L, and g(x) ϕ(l)[x] is obtained by applying ϕ to the coefficients of f(x), then K is also a splitting field of g(x) over ϕ(l). Then by our theorem on splitting fields, there exists some ψ Aut(K) which restricts to ϕ on L, as desired. We are now ready to prove our characterization of Galois extensions. Proof of Theorem 2.8. Given α K of degree d over F, let f(x) F [x] be the minimal polynomial of α over F, and suppose that f(x) has m roots in K, so that m d. We observe that there are exactly m inclusions of F (α) into K which fix F, since there is a unique one sending α to each root of f(x). We then conclude from Proposition 3.5 that Aut(K/F ) m Aut(K/F (α)), with equality if K/F (α) is a splitting field of some polynomial. (a) implies (c): If K/F is Galois, then for any α K as above, we have [K : F ] = Aut(K/F ) m Aut(K/F (α)) m[k : F (α)] = [K : F ] m/d, so we must have m = d, and we conclude that the minimal polynomial of α over F splits completely in K, and has distinct roots. (c) implies (b): If α 1,..., α n generate K over F, and f 1 (x),..., f n (x) are their minimal polynomials over F, then letting g(x) be the product of all distinct f i (x), we see that K is the splitting field of g(x) over F, and g(x) does not have multiple roots in K (note that if f i (x) and f j (x) had common roots in K, then they would have to have a nonconstant common factor also in F, contradicting irreducibility). (b) implies (a): The proof proceeds by induction on [K : F ], with the base case [K : F ] = 1 being trivially true. If [K : F ] > 1 and K/F is a splitting field of some polynomial g(x) F [x] which does not have multiple roots in K, let α be a root of g(x) not lying in F. Let f(x) be the minimal polynomial of α over F, and suppose f(x) has degree d. Then f(x) divides g(x), so f(x) must also split completely in K without multiple roots, meaning that f(x) has d roots in K. Observing that K is still a splitting field of g(x) over F (α), and [K : F (α)] < [K : F ], the induction hypothesis says that Aut(K/F (α)) = [K : F (α)]. From the above inequality, we conclude that Aut(K/F ) = d Aut(K/F (α)) = d[k : F (α)] = d[k : F ]/d = [K : F ], so K/F is Galois. For the case that F has characteristic 0, we need to show that if K/F is the splitting field of some polynomial f(x) F [x], then it is the splitting field of a polynomial g(x) F [x] which does not have multiple roots in K. But in characteristic 0, we showed in an earlier proposition that no irreducible polynomial has multiple roots in any extension field, so if we let g(x) be the product of the irreducible factors of f(x), then K/F is still a splitting field of g(x), and as above g(x) will not have any multiple roots in K. In order to prove Theorem 2.7, one key lemma (also due to Artin) remains: 7

8 Lemma 3.6. Let K be a field, and H Aut(K) a finite group of automorphisms. Then [K : K H ] H. Proof. The proof is very similar to those of Lemma 1.7 and Theorem 1.6, with the role of the field and the automorphisms switched. Let ϕ 1,..., ϕ n be the elements of H, ordered so that ϕ 1 is the identity, and suppose [K : K H ] > n. Then we can find elements β 1,..., β n+1 K which are linearly independent over K H. Consider the following system of n linear equations in n + 1 variables: ϕ 1 (β 1 )x 1 + ϕ 1 (β 2 )x ϕ 1 (β n+1 )x n+1 = 0 ϕ n (β 1 )x 1 + ϕ n (β 2 )x ϕ n (β n+1 )x n+1 = 0. By linear algebra, there exist (α 1,..., α n+1 ) K n+1 not all zero which solve the system of equations. Among all such tuples, consider one with a minimal number of nonzero entries. Because ϕ 1 = id, the first equation says that β 1 α β n+1 α n+1 = 0, and because the β i are linearly independent, they are in particular all nonzero, so we conclude that we cannot have exactly one α i nonzero. Without loss of generality, we may assume that α 1 and α n+1 are both nonzero, and dividing through by α n+1, we may actually assume α n+1 = 1. Since the β i are linearly independent over K H, not all the α i can be in K H, but α n+1 = 1, so reordering if necessary, we may assume that α 1 K H. Then by definition there exists i such that ϕ i (α 1 ) α 1. Applying ϕ i to both sides of each of the above equations, we find that (ϕ i (α 1 ),..., ϕ i (α n+1 )) is a solution to the system ϕ i ϕ 1 (β 1 )x 1 + ϕ i ϕ 1 (β 2 )x ϕ i ϕ 1 (β n+1 )x n+1 = 0 ϕ i ϕ n (β 1 )x 1 + ϕ i ϕ n (β 2 )x ϕ i ϕ n (β n+1 )x n+1 = 0. But because H is a group, the set {ϕ i ϕ 1,..., ϕ i ϕ n } is the same as the set {ϕ 1,..., ϕ n }, so the above system is the same as the original system, and (ϕ i (α 1 ),..., ϕ i (α n+1 )) is a solution to the original system as well. Now, α n+1 = 1, so ϕ i (α n+1 ) = 1 also. Moreover, for any j with α j = 0, we have ϕ i (α j ) = 0, so subtracting the two solutions, we have a solution with nonzero first entry, and strictly fewer nonzero terms than the original solution (α 1,..., α n+1 ). But this contradicts minimality. We now complete the proof of the main theorem of Galois theory. Proof of Theorem 2.7. For the first statement, it suffices to show that for all subgroups H Gal(K/F ), we have H = Gal(K/K H ), and for all intermediate fields L, we have L = K Gal(K/L). First, given a subgroup H, consider ϕ H. For any α K H, by definition ϕ(α) = α, so by definition ϕ Gal(K/K H ). That is, H Gal(K/K H ). On the other hand, Gal(K/K H ) = [K : K H ] H by Lemma 3.6, so we conclude that H must be all of Gal(K/K H ), as desired. Next, given L, and α L, then for all ϕ Gal(K/L), by definition ϕ(α) = α, so we conclude that α K Gal(K/L), and consequently that L K Gal(K/L). For the opposite inclusion, we want to show that if α K satisfies ϕ(α) = α for all α Gal(K/L), then α L. In fact, we will prove the contrapositive: if α L, we can find some ψ Gal(K/L) such that ψ(α) α. We know that K/L is Galois by Corollary 2.9, but if f(x) is the minimal polynomial of α over L, by Theorem 2.8 since f(x) has a root in K it must split completely in K, and must moreover not have multiple roots. In particular, there is a subfield E of K containing L such that E is the splitting field of f(x) over L. Since α L and f(x) doesn t have multiple roots, there exists some β α L which is also a root of f(x), and by our theorem on splitting fields there is an automorphism ϕ Aut(E/L) such that ϕ(α) = β. By Theorem 2.8, we have that K/L and hence K/E are splitting fields, so 8..

9 using splitting field theorem again, there exists ψ Aut(K/L) which is equal to ϕ on L, and then ψ(α) = β α, as desired. Now, suppose that H Gal(K/F ) is a normal subgroup. Observe that Gal(K/F ) = [K : F ], and [K H : F ] = [K : F ]/[K : K H ] = [K : F ]/ H, so in order to prove K H /F is Galois, we wish to show that Aut(K H /F ) = Gal(K/F ) / H. We will in fact show that that Aut(K H /F ) = Gal(K/F )/H. Given ϕ Gal(K/F ), we claim that for all α K H, we have ϕ(α) K H. Given ψ H, we observe that ϕ 1 ψϕ H by normality, so ϕ 1 ψϕ(α) = α, and applying ϕ to both sides, we get ψϕ(α) = ϕ(α). Since this holds for all ψ H, we get ϕ(α) K H, as claimed. Thus, every ϕ Gal(K/F ) restricts to give an element of Aut(K H /F ), and since restriction preserves composition, we have a group homomorphism Gal(K/F ) Aut(K H /F ). This homomorphism is surjective by our theorem on splitting fields, so it suffices to show the kernel is H. Certainly, H is contained in the kernel, but if ϕ Gal(K/F ) fixes K H, then ϕ Gal(K/K H ) = H, so the kernel is equal to H, as desired. Conversely, suppose that K H /F is Galois. Then by Theorem 2.8 we know that it is a splitting field, and then by Theorem 3.3, we know that for every ϕ Gal(K/F ), we have ϕ(k H ) K H. Given ψ H, consider ϕ 1 ψϕ. If α K H, then ϕ(α) K H as well, so ψϕ(α) = ϕ(α), and ϕ 1 ψϕ(α) = ϕ 1 ϕ(α) = α. Thus, ϕ 1 ψϕ Gal(K/K H ) = H, and since ϕ was arbitrary, we conclude H is normal, as desired. 9

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